Inheritance, Variation, and Evolution

Inheritance, Variation, and Evolution

:::info Board Coverage AQA Paper 2 | Edexcel Paper 2 | OCR A Gateway B3 | WJEC B3 :::

1. DNA and Genetics

1.1 DNA Structure

DNA (deoxyribonucleic acid) is a long molecule that carries the genetic code. It is found in the Nucleus of cells and is organised into structures called chromosomes.

Structure of DNA:

• Two strands forming a double helix (discovered by Watson and Crick in 1953, based on X-ray diffraction data from Rosalind Franklin)
• Each strand is made of nucleotides
• Each nucleotide contains: a sugar (deoxyribose), a phosphate group, and a base
• Four bases: adenine (A), thymine (T), cytosine (C), guanine (G)
• Base pairing: A always pairs with T (two hydrogen bonds); C always pairs with G (three hydrogen bonds)
• The two strands are antiparallel (they run in opposite directions)

Why base pairing is specific. The sizes and shapes of the bases determine which pairs can form. A and T form two hydrogen bonds because they have complementary shapes. C and G form three hydrogen Bonds. This specificity is fundamental to the accuracy of DNA replication: each base can only pair With its complementary partner, ensuring that the genetic information is copied correctly.

Nucleotide structure in detail. Each nucleotide consists of three parts:

1. A pentose sugar (deoxyribose) — a 5-carbon sugar.
2. A phosphate group — attached to the 5’ carbon of the sugar.
3. A nitrogenous base — attached to the 1’ carbon of the sugar.

Nucleotides are joined together by phosphodiester bonds between the phosphate group of one Nucleotide and the sugar of the next, forming a sugar-phosphate backbone. The bases project inwards From the backbone and pair with complementary bases on the opposite strand.

1.2 Genes and Chromosomes

• A gene is a short section of DNA on a chromosome that codes for a specific protein (and therefore a specific characteristic). The sequence of bases in a gene determines the sequence of amino acids in the protein, which determines the protein’s shape and function.
• Humans have 23 pairs of chromosomes (46 total). One of each pair comes from the mother and one from the father.
• Each chromosome contains many genes (approximately 20,000—25,000 genes in the human genome).
• The complete set of genes in an organism is called the genome.

Worked Example: Calculating the number of bases in DNA.

The human haploid genome contains approximately 3 billion base pairs. Since DNA is double-stranded, The total number of bases in one diploid cell is:

$3 \times 10^9 \times 2 = 6 \times 10^9$ bases (6 billion bases per diploid cell).

If the average distance between bases is 0.34 nm, the total length of DNA in one cell is:

6 \times 10^9 \times 0.34 \times 10^{-9} \mathrm{ m = 2.04 \mathrm{ m.

This 2 metres of DNA is packed into a nucleus that is only about 10 micrometres in diameter, Demonstrating the remarkable efficiency of DNA packaging.

1.3 Protein Synthesis

The genetic code in DNA is used to make proteins through a two-stage process:

1. Transcription: The DNA code is copied into mRNA (messenger RNA) in the nucleus. RNA polymerase binds to the DNA and synthesises a complementary mRNA strand. The mRNA is a “working copy” of the gene that can leave the nucleus.
2. Translation: The mRNA moves to a ribosome, where its code is read three bases at a time (each group of three bases is called a codon). Each codon specifies a particular amino acid. Transfer RNA (tRNA) molecules bring the correct amino acids, and they are joined together by peptide bonds to form a polypeptide chain. The chain folds into its functional protein shape.

The sequence of bases in a gene determines the sequence of amino acids in a protein, which Determines the protein’s shape and function. A change in the DNA sequence (mutation) can change the Amino acid sequence, potentially altering the protein’s shape and function.

Summary of protein synthesis steps.

Stage	Location	Template	Product	Key molecules involved
Transcription	Nucleus	DNA	mRNA	RNA polymerase
Translation	Ribosome	mRNA	Protein	tRNA, ribosomes, amino acids

1.4 Genome

The genome is the entire genetic material of an organism. The Human Genome Project (completed In 2003) mapped all the genes in human DNA.

Benefits of the Human Genome Project:

• Understanding genetic diseases and developing new treatments (e.g., identifying the mutations that cause cystic fibrosis or Huntington’s disease)
• Predicting disease risk (identifying alleles associated with increased risk of certain cancers or heart disease)
• Tracing human migration patterns (comparing DNA sequences from populations around the world)
• Developing personalised medicine (tailoring drug treatments to an individual’s genetic profile)
• Forensic applications (DNA fingerprinting for identifying individuals)

Concerns about genome data:

• Privacy: who should have access to an individual’s genetic information?
• Genetic discrimination: could employers or insurance companies use genetic data to make decisions?
• Psychological impact: knowing you carry a gene for a serious disease can cause anxiety.

1.5 Higher Tier: The Genetic Code

The genetic code has three important properties:

• Degenerate: Most amino acids are coded for by more than one codon. For example, the amino acid leucine is coded by six different codons. This provides some protection against mutations: if one base changes, the same amino acid may still be specified.
• Universal: The same codons code for the same amino acids in almost all organisms (from bacteria to humans). This is powerful evidence for common ancestry and allows genes from one organism to be expressed in another (the basis of genetic engineering).
• Non-overlapping: Each base is part of only one codon. The code is read in a fixed reading frame from a specific start point.

2. Reproduction

2.1 Sexual Reproduction

Sexual reproduction involves the fusion of two gametes (sex cells: sperm and egg) to form a zygote. The offspring inherits genetic information from both parents, producing genetic Variation.

Meiosis is the type of cell division that produces gametes:

1. The cell replicates its DNA (each chromosome has two identical chromatids).
2. The cell divides twice:

• First division: Homologous chromosomes separate (reducing the chromosome number by half). Crossing over occurs during Prophase I, where non-sister chromatids exchange segments of DNA.
• Second division: Chromatids separate (similar to mitosis).

3. Four genetically unique daughter cells are produced, each with half the normal number of chromosomes (haploid).

Key differences from mitosis:

Feature	Mitosis	Meiosis
Divisions	One	Two
Daughter cells	2 (identical)	4 (genetically different)
Chromosome number	Same as parent	Half of parent
Purpose	Growth, repair	Producing gametes
Crossing over?	No	Yes (Prophase I)

2.2 Asexual Reproduction

Asexual reproduction involves only one parent. The parent cell divides by mitosis to produce Genetically identical offspring (clones).

Examples: Bacteria (binary fission), strawberry runners, potato tubers, spider plants.

Advantages:

• Only one parent needed (no need to find a mate)
• Rapid (many offspring quickly)
• All offspring are well-adapted if the environment is stable

Disadvantages:

• No genetic variation (vulnerable to disease and environmental change; if a disease kills one individual, it can kill all of them)
• Overcrowding and competition for resources

Worked Example: Bacterial growth by binary fission.

A single bacterium divides by binary fission every 20 minutes. How many bacteria will there be after 4 hours?

Step 1: Calculate the number of divisions.

4 hours = 240 minutes. $240 / 20 = 12$ divisions.

Step 2: Calculate the number of bacteria.

After $n$ divisions, the number of bacteria is $2^n$.

$2^{12} = 4096$ bacteria.

This shows how rapidly asexual reproduction can produce large numbers of organisms.

2.3 DNA Extraction (Required Practical)

Method:

1. Mash strawberries (or kiwi) in a zip-lock bag with salt water and detergent.

• Salt: neutralises the negative charges on DNA, allowing it to clump together.
• Detergent: breaks down the cell membranes and nuclear membranes, releasing the DNA.

2. Filter the mixture to remove the pulp (cell debris, proteins, and other large molecules).
3. Add cold ethanol to the filtrate.

• DNA is insoluble in cold ethanol, so it precipitates out of solution.

4. DNA precipitates out as a white, stringy substance that can be spooled onto a glass rod.

2.4 Higher Tier: Sexual and Asexual Reproduction Compared

Feature	Sexual Reproduction	Asexual Reproduction
Number of parents	Two	One
Gametes involved?	Yes	No
Cell division	Meiosis + fertilisation	Mitosis
Genetic variation	High (offspring are unique)	None (offspring are clones)
Speed	Slower (need to find a mate)	Faster
Adaptability	High (variety aids adaptation)	Low (all vulnerable to same threats)

3. Inheritance

3.1 Key Terms

Term	Definition
Gene	A section of DNA that codes for a specific protein
Allele	A version of a gene (e.g. Brown or blue eye colour)
Dominant	An allele that is expressed when one or two copies are present (capital letter, e.g. B)
Recessive	An allele that is expressed only when two copies are present (lowercase letter, e.g. B)
Homozygous	Two identical alleles for a gene (e.g. BB or bb)
Heterozygous	Two different alleles for a gene (e.g. Bb)
Genotype	The combination of alleles an individual has (e.g. Bb)
Phenotype	The observable characteristic produced by the genotype (e.g. Brown eyes)
Carrier	A heterozygous individual who carries a recessive allele without showing the characteristic

3.2 Genetic Crosses

Monohybrid crosses investigate the inheritance of one characteristic.

Worked Example 1. In pea plants, tall (T) is dominant over dwarf (t). Cross a heterozygous tall Plant with a dwarf plant.

Parental genotypes: Tt $\times$ tt

Punnett square:

	T	t
t	Tt	tt
t	Tt	tt

Genotypic ratio: 2 Tt : 2 tt (1:1)

Phenotypic ratio: 2 tall : 2 dwarf (1:1)

Probability of tall offspring: 50%

Worked Example 2. Cystic fibrosis is caused by a recessive allele (f). Both parents are carriers (Ff).

Punnett square:

	F	f
F	FF	Ff
f	Ff	ff

Genotypic ratio: 1 FF : 2 Ff : 1 ff

Phenotypic ratio: 3 healthy : 1 cystic fibrosis

Probability of a child with cystic fibrosis: 25%

Understanding why carriers do not show the disease. A carrier (Ff) has one dominant allele (F) That produces enough functional protein to maintain normal health. The recessive allele (f) does not Produce functional protein, but one copy of the dominant allele is sufficient. This is the principle Of dominance: the dominant allele masks the effect of the recessive allele in the heterozygote.

Worked Example 3: A two-generation cross.

In guinea pigs, black fur (B) is dominant over white fur (b). A homozygous black guinea pig is Crossed with a white guinea pig.

F0 cross: BB $\times$ bb

All F1 offspring are Bb (heterozygous black).

Now cross two F1 offspring: Bb $\times$ Bb

Punnett square:

	B	b
B	BB	Bb
b	Bb	bb

F2 genotypic ratio: 1 BB : 2 Bb : 1 bb

F2 phenotypic ratio: 3 black : 1 white

Probability of a white guinea pig in the F2 generation: 25%.

3.3 Sex Determination

Sex is determined by the X and Y chromosomes:

• Females: XX
• Males: XY

Sex of offspring:

	X	X
X	XX (female)	XX (female)
Y	XY (male)	XY (male)

Probability of male or female: 50% each. This is because the father can contribute either an X or a Y chromosome, while the mother always contributes an X chromosome.

3.4 Inherited Disorders

Disorder	Pattern of Inheritance
Cystic fibrosis	Recessive (caused by mutation in CFTR gene)
Huntington’s disease	Dominant (caused by mutation on chromosome 4)
Sickle cell disease	Recessive (caused by mutation in haemoglobin gene)
Polydactyly	Dominant (extra fingers or toes)

Why recessive disorders are more common than dominant disorders. A recessive disorder can be “hidden” in carriers (heterozygotes) who do not show the disease. These carriers can pass the Recessive allele to their children without knowing it. A dominant disorder cannot be hidden: anyone With the allele shows the disease. Dominant disorders are therefore more likely to be selected Against (affected individuals may have reduced fitness and fewer children), while recessive Disorders can persist in the population for many generations through carriers.

Genetic testing and screening:

• Pre-implantation genetic diagnosis (PGD): Embryos are tested before implantation during IVF. Only embryos without the disorder are implanted.
• Antenatal testing: Testing the fetus during pregnancy (e.g., amniocentesis, chorionic villus sampling).
• Genetic counselling: Advising families about the risks of inherited conditions.

Ethical considerations:

• Right to know vs. Right not to know (should parents be told if their unborn child has a genetic disorder?)
• Potential for discrimination based on genetic information (by employers or insurance companies)
• Implications for insurance and employment (genetic discrimination)
• Decisions about termination of pregnancy (is it ethical to terminate a pregnancy because the fetus has a genetic disorder?)

3.5 Higher Tier: Sex-Linked Inheritance

Some genes are located on the X chromosome. Because males have only one X chromosome (XY), a single Recessive allele on the X chromosome will be expressed in males. Females (XX) need two copies of the Recessive allele to show the condition.

Example: Red-green colour blindness is caused by a recessive allele on the X chromosome ($X^n$). A carrier female ($X^NX^n$) crossed with a normal male ($X^NY$):

	$X^N$	$Y$
$X^N$	$X^NX^N$ (normal female)	$X^NY$ (normal male)
$X^n$	$X^NX^n$ (carrier female)	$X^nY$ (affected male)

This cross shows that colour blindness affects males more frequently than females. In this cross, 50% of males are affected but no females are affected (though 50% are carriers).

Worked Example: Haemophilia inheritance.

Haemophilia is an X-linked recessive disorder. A woman who is a carrier ($X^HX^h$) has children with A man who does not have haemophilia ($X^HY$).

Punnett square:

	$X^H$	$Y$
$X^H$	$X^HX^H$ (normal female)	$X^HY$ (normal male)
$X^h$	$X^HX^h$ (carrier female)	$X^hY$ (haemophiliac)

Results: 25% normal female, 25% carrier female, 25% normal male, 25% haemophiliac male.

The probability that a son will have haemophilia: 50% (because a son inherits his only X chromosome From his mother, and there is a 50% chance she passes on the $X^h$ allele).

4. Variation and Evolution

4.1 Sources of Variation

Genetic variation is caused by:

• Sexual reproduction (combination of alleles from two parents)
• Mutation (changes in DNA sequence — the ultimate source of all new variation)
• Meiosis (independent assortment and crossing over create new combinations of alleles)

Environmental variation is caused by:

• Diet (affecting height, weight, and health)
• Climate (affecting skin colour through tanning)
• Lifestyle (exercise affecting muscle mass, smoking affecting lung function)
• Accidents (scars, amputations)

Most characteristics are determined by a combination of genetic and environmental factors (e.g. Height, weight, intelligence). Height, for example, is largely determined by genetics (many genes Contribute) but is also influenced by nutrition during childhood.

Worked Example: Classifying types of variation.

Classify each characteristic as genetic (G), environmental (E), or both (B):

• Eye colour: G (determined by genes)
• Height: B (genes determine potential, nutrition affects actual height)
• Scars: E (caused by injury)
• Blood group: G (determined by genes)
• Skin colour from tanning: E (caused by UV exposure)
• Body mass: B (genes influence metabolism, diet and exercise affect actual mass)
• Hair colour: G (determined by genes, though it can change with age)
• Accent: E (determined by environment)

4.2 Evolution by Natural Selection

Darwin’s theory of natural selection:

1. Individuals within a species show variation (due to genetic differences).
2. There is competition for limited resources (food, mates, territory).
3. Individuals with characteristics better suited to the environment are more likely to survive and reproduce.
4. They pass on their advantageous alleles to their offspring.
5. Over many generations, the frequency of advantageous alleles increases in the population.
6. The species evolves.

Key insight: Natural selection does not act on genes directly; it acts on phenotypes (observable Characteristics). The phenotype that confers a survival or reproductive advantage is selected for, But the alleles that produce that phenotype are what are passed to the next generation.

Example: Peppered moths during the Industrial Revolution.

• Before industrialisation: Light moths were better camouflaged on lichen-covered trees, so they were eaten less frequently by birds. The frequency of the light allele was high.
• During industrialisation: Soot darkened tree bark, killing the lichen. Dark moths became better camouflaged. The frequency of the dark allele increased through natural selection.
• After clean-air legislation: Tree bark became lighter again. The frequency of the light allele began to increase.

This example demonstrates that natural selection is not directional in a fixed sense; it responds to Changes in the environment.

4.3 Evidence for Evolution

Evidence	Description
Fossil record	Shows gradual changes in organisms over millions of years
DNA comparison	More similar DNA = more closely related species
Anatomy	Homologous structures suggest common ancestry (e.g. Pentadactyl limb)
Embryology	Early embryos of different species look remarkably similar

Homologous vs. Analogous structures. Homologous structures have the same evolutionary origin but May serve different functions (e.g., the pentadactyl limb in humans, bats, whales, and birds — all Have the same basic bone structure despite being used for grasping, flying, swimming, and walking Respectively). Analogous structures have different evolutionary origins but serve similar functions (e.g., the wings of birds and insects). Only homologous structures provide evidence for common Ancestry.

4.4 Speciation

Speciation is the formation of a new species. It occurs when populations of the same species Become reproductively isolated and evolve differently.

Mechanism (allopatric speciation):

1. A geographical barrier separates a population (e.g., mountain range, river, ocean).
2. The two populations experience different environmental conditions and selection pressures.
3. Over time, each population accumulates different mutations and adaptations.
4. Eventually, the two populations become so different that they can no longer interbreed to produce fertile offspring.
5. A new species has formed.

Why reproductive isolation is the defining criterion. Two populations may look different, but if They can still interbreed and produce fertile offspring, they are considered the same species. Speciation is complete only when gene flow between the populations ceases entirely.

4.5 Antibiotic Resistance

Bacteria can develop resistance to antibiotics through natural selection:

1. A mutation occurs in a bacterium, giving it resistance to an antibiotic.
2. When the antibiotic is used, non-resistant bacteria are killed.
3. The resistant bacterium survives and reproduces.
4. The resistance allele spreads through the population.

Why antibiotic resistance is a major public health concern. The rate at which bacteria evolve Resistance is outpacing the rate at which new antibiotics are being developed. If antibiotics lose Their effectiveness, routine medical procedures (surgery, organ transplants, cancer treatment) Become much more dangerous because the risk of untreatable infections increases.

Preventing antibiotic resistance:

• Only use antibiotics when necessary (not for viral infections like the common cold, which antibiotics cannot treat)
• Complete the full course of antibiotics (stopping early allows partially resistant bacteria to survive)
• Develop new antibiotics (research into new classes of antibiotics is urgently needed)
• Reduce antibiotic use in agriculture (antibiotics are widely used in livestock farming, which contributes to the spread of resistance)

4.6 Classification

Organisms are classified into a hierarchy:

Domain $\to$ Kingdom $\to$ Phylum $\to$ Class $\to$ Order $\to$ Family $\to$ Genus $\to$ Species

The binomial naming system (Carl Linnaeus) gives each species a two-part Latin name:

• First part: Genus (capitalised)
• Second part: Species (lowercase)
• Example: Homo sapiens

Carl Woese proposed three domains based on RNA analysis:

• Bacteria
• Archaea
• Eukarya (includes animals, plants, fungi, protists)

5. Higher Tier: Selective Breeding and Genetic Engineering

5.1 Selective Breeding

Selective breeding (artificial selection) is the process by which humans choose organisms with Desirable characteristics to be parents of the next generation.

Examples:

• Cattle bred for increased milk yield or lean meat.
• Wheat bred for disease resistance or higher protein content.
• Dogs bred for specific physical or behavioural traits (resulting in the many different breeds we see today).

Process:

1. Identify individuals with the desired characteristic.
2. Breed them together.
3. Select the offspring with the best expression of the characteristic.
4. Repeat over many generations.

Advantages: Can produce organisms with desirable traits (higher yield, disease resistance).

Disadvantages: Reduces genetic variation (all individuals become genetically similar, making the Population vulnerable to disease). Can cause health problems in animals (e.g., hip dysplasia in some Dog breeds, breathing difficulties in pugs).

5.2 Genetic Engineering

Genetic engineering involves transferring genes from one organism to another to give it a desired Characteristic.

Process:

1. The desired gene is identified and cut from the DNA using restriction enzymes.
2. The gene is inserted into a vector (e.g., a bacterial plasmid) using DNA ligase.
3. The vector is introduced into the host organism (e.g., a bacterium).
4. The host organism expresses the gene and produces the desired protein.

Examples:

• Human insulin gene inserted into bacteria for mass production of insulin (used to treat type 1 diabetes). This replaced the previous method of extracting insulin from pig pancreases.
• GM crops: insect-resistant maize (contains a gene from Bacillus thuringiensis that produces a toxin lethal to insect pests), herbicide-resistant soya beans.

Advantages: Can produce organisms with specific, useful traits. Medical applications (insulin, Growth hormone, clotting factors).

Disadvantages and ethical concerns:

• Potential ecological risks (GM crops could cross-breed with wild relatives, spreading the modified genes into natural ecosystems).
• Long-term health effects of GM foods are not fully known.
• Ethical concerns about modifying the genetic code of living organisms.
• Patents on GM seeds can make farmers dependent on large biotechnology companies.

Comparison of selective breeding and genetic engineering.

Feature	Selective Breeding	Genetic Engineering
Process	Breeding organisms with desired traits	Transferring specific genes between organisms
Speed	Slow (many generations)	Fast (one generation)
Specificity	Cannot select single genes	Can select specific genes
Genetic variation	Reduced over time	Can introduce genes from any species
Ethical concerns	Animal welfare	Unnatural gene transfer, ecological risks

Common Pitfalls

• Confusing mitosis and meiosis. Mitosis: 2 identical cells (growth/repair). Meiosis: 4 unique cells (gametes). Crossing over only occurs in meiosis.
• Using the wrong notation for alleles. Dominant = capital letter (B); recessive = lowercase (b). Never use different letters for the same gene.
• Forgetting that carriers are heterozygous. A carrier has one dominant and one recessive allele (Bb) and does not show the recessive characteristic. Only recessive homozygotes (bb) show the recessive phenotype.
• Confusing genotype and phenotype. Genotype = alleles (e.g. Bb); phenotype = physical appearance (e.g. Brown eyes). The genotype determines the phenotype.
• Thinking evolution happens to individuals. Evolution occurs in populations over many generations, not within an individual’s lifetime. An individual cannot evolve; a population evolves as allele frequencies change.
• Confusing Darwin’s theory with Lamarck’s. Lamarck proposed inheritance of acquired characteristics (e.g. Giraffes stretch their necks and pass this on) — this is incorrect. Darwin proposed natural selection (giraffes with longer necks survive better and pass on the genes for longer necks).
• Forgetting that dominant does not mean more common. A dominant allele is expressed when present, but it may be rare in the population (e.g., the allele for Huntington’s disease is dominant but rare).
• Confusing the terms “gene” and “allele.” A gene is a section of DNA coding for a protein. Alleles are different versions of the same gene.
• Writing probabilities as percentages incorrectly. A 3:1 ratio means 75% dominant and 25% recessive, not 3% and 1%.
• Forgetting to state the expected ratio before the Punnett square. Always show the parental genotypes and the gametes .

Practice Questions

1. Describe the structure of DNA, including how the bases pair.

2. Explain the difference between mitosis and meiosis, including when each type of cell division occurs.

3. In cats, the allele for black fur (B) is dominant over the allele for white fur (b). A heterozygous black cat mates with a white cat. Use a Punnett square to show the possible genotypes and phenotypes of the offspring.

4. Explain how antibiotic resistance develops in bacteria, with reference to natural selection.

5. Describe the process of speciation, using an example.

6. Explain why sexual reproduction produces genetic variation but asexual reproduction does not.

7. Describe two types of evidence that support the theory of evolution.

8. Both parents are carriers of cystic fibrosis (Ff). Calculate the probability that their child will have cystic fibrosis.

9. Explain the difference between genetic and environmental variation, giving examples of each.

10. Describe the binomial naming system and explain why it is useful for scientists worldwide.

11. (Higher Tier) Explain why males are more likely to show sex-linked recessive conditions than females, using a genetic cross to support your answer.

12. (Higher Tier) Describe the process of selective breeding and explain two potential disadvantages.

13. A genetic cross between two pea plants produces offspring in a ratio of 3 tall : 1 dwarf. Explain this result, naming the probable genotypes of the parents.

14. Explain the role of mutation in evolution. Why are most mutations harmful, but some beneficial?

15. Describe how genetic engineering could be used to produce human insulin, and discuss one advantage and one disadvantage of this approach.

16. A woman with blood group A (genotype $I^A i$) has a child with a man who has blood group B (genotype $I^B i$). What are the possible blood groups of their child? Use a Punnett square to show your working.

17. Explain the difference between homologous and analogous structures, giving one example of each.

18. Describe the process of natural selection using antibiotic resistance in bacteria as an example.

19. A farmer wants to breed chickens that lay more eggs. Describe the process of selective breeding they would use and explain one potential disadvantage of this approach.

20. Explain why the Human Genome Project is both beneficial and controversial, giving at least two examples of each.

6. Higher Tier: Codominance and Incomplete Dominance

Not all inheritance follows simple dominant/recessive patterns. Two important exceptions are Codominance and incomplete dominance.

6.1 Codominance

In codominance, both alleles are expressed fully in the heterozygote. Neither allele is dominant or Recessive. The heterozygote shows both phenotypes simultaneously.

Example: Blood groups. Human ABO blood groups are determined by three alleles: $I^A$, $I^B$And $i$. $I^A$ and $I^B$ are codominant to each other, and both are dominant over $i$.

Genotype	Blood group	Phenotype
$I^A I^A$	A	Antigen A on red blood cells
$I^A i$	A	Antigen A on red blood cells
$I^B I^B$	B	Antigen B on red blood cells
$I^B i$	B	Antigen B on red blood cells
$I^A I^B$	AB	Both antigens A and B (codominant)
$ii$	O	Neither antigen

Worked Example: Blood group inheritance.

A mother has blood group A (genotype $I^A i$) and a father has blood group B (genotype $I^B i$). What are the possible blood groups of their children?

Punnett square:

	$I^A$	$i$
$I^B$	$I^A I^B$ (AB)	$I^B i$ (B)
$i$	$I^A i$ (A)	$ii$ (O)

Possible blood groups: A, B, AB, and O (all four groups are possible, each with 25% probability).

6.2 Incomplete Dominance

In incomplete dominance, the heterozygote shows an intermediate phenotype between the two Homozygotes. The alleles blend rather than showing both fully.

Example: In snapdragons, the allele for red flowers (R) is incompletely dominant over the allele For white flowers (W). The heterozygote (RW) has pink flowers.

Cross: RR (red) $\times$ WW (white) $\to$ all RW (pink).

Cross: RW (pink) $\times$ RW (pink):

	R	W
R	RR (red)	RW (pink)
W	RW (pink)	WW (white)

Phenotypic ratio: 1 red : 2 pink : 1 white.

6.3 Multiple Alleles

Some genes have more than two alleles in the population, although any individual can only carry two Alleles (one on each chromosome). The ABO blood group system is an example: there are three alleles ($I^A$, $I^B$And $i$), but each person carries only two of them.

7. Higher Tier: Pedigree Charts

A pedigree chart is a diagram showing the inheritance of a particular trait through several Generations of a family. Pedigree charts are used by genetic counsellors to determine the pattern of Inheritance of a genetic disorder and to calculate the probability of future offspring being Affected.

Conventions:

• Males are represented by squares, females by circles.
• Shaded symbols indicate affected individuals.
• Unshaded symbols indicate unaffected individuals.
• Horizontal lines connect mates; vertical lines connect parents to offspring.
• Carriers (heterozygous for a recessive allele) are sometimes shown as half-shaded.

How to determine the inheritance pattern from a pedigree:

1. Autosomal recessive: Appears in both sexes equally; affected children can have unaffected parents (carriers); skips generations; two affected parents always have affected children.
2. Autosomal dominant: Appears in both sexes equally; affected children always have at least one affected parent; does not skip generations.
3. X-linked recessive: More males affected than females; affected males pass the allele to all daughters (who become carriers) but not to sons; cannot be passed from father to son.

Worked Example: Interpreting a pedigree chart.

A pedigree chart for a rare disorder shows:

• Both males and females are affected in approximately equal numbers.
• Two unaffected parents have an affected child.
• The disorder skips a generation (grandparents are unaffected but a grandchild is affected).

Conclusion: This disorder follows an autosomal recessive pattern. The evidence is:

• Equal numbers of affected males and females $\to$ not X-linked.
• Unaffected parents having an affected child $\to$ must be recessive (both parents are carriers).
• Skipping generations $\to$ consistent with a recessive allele being hidden in heterozygous carriers.

8. Higher Tier: Genetic Crosses with Multiple Characteristics

A dihybrid cross investigates the inheritance of two characteristics simultaneously. Mendel’s Law of independent assortment states that alleles of different genes are distributed independently Of one another during gamete formation (provided the genes are on different chromosomes).

Worked Example: Dihybrid cross in pea plants.

In pea plants, round seeds (R) are dominant over wrinkled seeds (r), and yellow seeds (Y) are Dominant over green seeds (y). Both genes are on different chromosomes. Cross two double Heterozygotes: RrYy $\times$ RrYy.

Each parent can produce four types of gamete: RY, Ry, rY, ry (in equal proportions).

Punnett square (4 $\times$ 4):

	RY	Ry	rY	ry
RY	RRYY (RY)	RRYy (RY)	RrYY (RY)	RrYy (RY)
Ry	RRYy (RY)	RRyy (Ry)	RrYy (RY)	Rryy (Ry)
rY	RrYY (RY)	RrYy (RY)	rrYY (rY)	rrYy (rY)
ry	RrYy (RY)	Rryy (Ry)	rrYy (rY)	rryy (ry)

Phenotypic ratio: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

This 9:3:3:1 ratio is characteristic of a dihybrid cross involving two independently assorting Genes, each with a dominant and recessive allele.

Proof that the ratio arises from two independent 3:1 ratios:

• Probability of round: 3/4. Probability of yellow: 3/4. Round and yellow: 3/4 $\times$ 3/4 = 9/16.
• Probability of round: 3/4. Probability of green: 1/4. Round and green: 3/4 $\times$ 1/4 = 3/16.
• Probability of wrinkled: 1/4. Probability of yellow: 3/4. Wrinkled and yellow: 1/4 $\times$ 3/4 = 3/16.
• Probability of wrinkled: 1/4. Probability of green: 1/4. Wrinkled and green: 1/4 $\times$ 1/4 = 1/16.

This confirms the 9:3:3:1 ratio.

Practice Problems

Question 1: Monohybrid cross

In pea plants, tall stems (T) are dominant over short stems (t). A heterozygous tall plant is crossed with a short plant. What are the expected genotype and phenotype ratios of the offspring?

Answer

Parental cross: $Tt \times tt$.

Gametes: T$$t and t$$t.

Punnett square:

	T	t
t	Tt	tt
t	Tt	tt

Genotype ratio: 1 Tt : 1 tt (50% : 50%). Phenotype ratio: 1 tall : 1 short (50% : 50%).

Question 2: Sex-linked inheritance

Red-green colour blindness is caused by a recessive allele on the X chromosome. A carrier female ($X^NX^n$) and a normal male ($X^NY$) have children. What is the probability that their son will be colour blind?

Answer

Punnett square:

	$X^N$	Y
$X^N$	$X^NX^N$	$X^NY$
$X^n$	$X^NX^n$	$X^nY$

Each child has a 50% chance of being a boy. For sons: 50% normal ($X^NY$), 50% colour blind ($X^nY$).

Probability of a colour blind son = P(son) $\times$ P(colour blind | son) = $0.5 \times 0.5 = 0.25$ (25%).

Question 3: DNA and protein synthesis

Describe the process of transcription. Explain how mRNA is used to produce a protein at the ribosome.

Answer

Transcription: The enzyme RNA polymerase binds to the DNA at the start of a gene. It unwinds the double helix and uses one strand as a template to build a complementary mRNA molecule. The base pairing rules are A-U (instead of A-T) and C-G. Once complete, the mRNA detaches and leaves the nucleus through a nuclear pore.

Translation (protein synthesis): The mRNA attaches to a ribosome. Transfer RNA (tRNA) molecules bring amino acids to the ribosome, matching their anticodon to the codon on the mRNA. The ribosome joins the amino acids together with peptide bonds, forming a polypeptide chain. The chain folds into a functional protein.

Question 4: Natural selection and antibiotic resistance

Explain how the overuse of antibiotics has led to the development of antibiotic-resistant bacteria, using the concept of natural selection.

Answer

In any bacterial population, random mutations occur. Some bacteria may acquire a mutation that provides resistance to an antibiotic. When antibiotics are used, susceptible bacteria are killed, but resistant bacteria survive. These survivors reproduce, passing the resistance allele to their offspring. Over time, the proportion of resistant bacteria in the population increases. Overuse of antibiotics creates a strong selection pressure that favours resistant strains. This is natural selection in action: the antibiotic acts as the selective pressure, and resistance is the advantageous trait.

Question 5: Genetic variation

Describe two sources of genetic variation and explain why genetic variation is important for the survival of a species.

Answer

1. Mutation: a random change in the DNA sequence that can create new alleles.
2. Sexual reproduction: meiosis produces gametes with different combinations of alleles through independent assortment and crossing over. Random fertilisation further increases variation.

Genetic variation is important because it provides the raw material for natural selection. If the environment changes (e.g., new disease, climate change), some individuals may possess advantageous alleles that allow them to survive and reproduce. Without variation, a species cannot adapt and is more vulnerable to extinction.

Worked Examples

Example 1:

A typical exam question on Inheritance, Variation, and Evolution requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.

Example 2:

Multi-step problems in Inheritance, Variation, and Evolution often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.

Summary

This topic covers the biological principles of inheritance, variation, and evolution, including key concepts, experimental evidence, and real-world applications.

Key concepts include:

• Mendelian inheritance
• gene expression and regulation
• mutations and genetic variation
• genetic engineering (PCR, gel electrophoresis)
• genome projects

Success requires the ability to recall specific factual content, apply knowledge to novel scenarios, and evaluate experimental evidence critically.